Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 965: 77

Answer

The function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.

Work Step by Step

Here, we have $u_x=-\dfrac{x}{(x^2+y^2+z^2)^{3/2}}\\u_{xx}=\dfrac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(A) and $u_{yy}=\dfrac{(2y^2-x^2-z^2)}{(x^2+y^2+z^2)^{5/2}}$ ...(B) Also, $u_{zz}=\dfrac{(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}$ ...(C) We can see from the equations (A), (B) and (C) that $u_{xx}+u_{yy}+u_{zz}=\dfrac{(2x^2-y^2-z^2)+(2y^2-x^2-z^2)+(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}=0$ Thus, it has been verified that the function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.
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